Rd sharma solution class 12 chapter Inverse trigonometric functions

Exercise 4.1

Question 1 

Find the principal value of each of the following:

(i) sin^{-1} \left(-\dfrac{\sqrt{3}}{~~2}\right)

Sol :

=sin^{-1}\left[sin~\left(-\dfrac{\pi}{3}\right)\right]

=-\dfrac{\pi}{3}

 

(ii) sin^{-1} \left(cos~\dfrac{2\pi}{3}\right)

Sol :

=sin^{-1} \left(-\dfrac{1}{2}\right)

=sin^{-1} \left[sin~\left(-\dfrac{\pi}{6}\right)\right]

=-\dfrac{\pi}{6}

 

(iii) sin^{-1} \left(\dfrac{\sqrt{3}-1}{2\sqrt{2}}\right)

Sol :

=sin^{-1}\left(\dfrac{\sqrt{3}}{2\sqrt{2}}-\dfrac{1}{2\sqrt{2}}\right)

=sin^{-1}\left(\dfrac{\sqrt{3}}{2}\times\dfrac{1}{\sqrt{2}}-\dfrac{1}{2}\times\dfrac{1}{\sqrt{2}}\right)

=sin^{-1}\left[\dfrac{\sqrt{3}}{2}\times \sqrt{1-\left(\dfrac{1}{\sqrt{2}}\right)^2}-\dfrac{1}{2}\times\sqrt{1-\left(\dfrac{\sqrt{3}}{2}\right)^2}\right]    \bigg[\because sin^{-1}~x-sin^{-1}~y=sin^{-1}\left(x~\sqrt{1-\left(y\right)^2}-y~\sqrt{1-\left(x\right)^2}\right)\bigg]

=sin^{-1}~\left(\dfrac{\sqrt{3}}{2}\right)-sin^{-1}~\left(\dfrac{1}{\sqrt{2}}\right)

=\dfrac{\pi}{3}-\dfrac{\pi}{4}

=\dfrac{\pi}{12}

 

(iv) sin^{-1} \left(\dfrac{\sqrt{3}+1}{2\sqrt{2}}\right) 

Sol :

=sin^{-1}\left(\dfrac{\sqrt{3}}{2\sqrt{2}}+\dfrac{1}{2\sqrt{2}}\right)

=sin^{-1}\left(\dfrac{\sqrt{3}}{2}\times\dfrac{1}{\sqrt{2}}+\dfrac{1}{2}\times\dfrac{1}{\sqrt{2}}\right)

=sin^{-1}\left[\dfrac{\sqrt{3}}{2}\times \sqrt{1-\left(\dfrac{1}{\sqrt{2}}\right)^2}+\dfrac{1}{2}\times\sqrt{1-\left(\dfrac{\sqrt{3}}{2}\right)^2}\right]          \bigg[\because sin^{-1}~x+sin^{-1}~y=sin^{-1}\left(x~\sqrt{1-\left(y\right)^2}+y~\sqrt{1-\left(x\right)^2}\right)\bigg]

=sin^{-1}~\left(\dfrac{\sqrt{3}}{2}\right)+sin^{-1}~\left(\dfrac{1}{\sqrt{2}}\right)

=\dfrac{\pi}{3}+\dfrac{\pi}{4}

=\dfrac{7\pi}{12}

 

(v) sin^{-1} \left(cos~\dfrac{3\pi}{4}\right)

Sol :

=sin^{-1} \left(-\dfrac{1}{\sqrt{2}}\right)

=sin^{-1}\left[sin \left(-\dfrac{\pi}{4}\right)\right]

=-\dfrac{\pi}{4}

 

(vi) sin^{-1} \left(tan~\dfrac{5\pi}{4}\right)

Sol :

Let y=sin^{-1}\left(tan\dfrac{5\pi}{4}\right)

Therefore sin~y=\left(tan\dfrac{5\pi}{4}\right) =tan\left(\pi+\dfrac{\pi}{4}\right) =tan\left(\dfrac{\pi}{4}\right) =1=sin\dfrac{\pi}{2}

We know that principal value of sin^{-1}\text{ is }\left[\dfrac{\pi}{2},\dfrac{\pi}{2}\right] and sin\left(\dfrac{\pi}{2}\right)=tan\left(\dfrac{5\pi}{4}\right)

Therefore principle value of sin^{-1}\left(tan\left(\dfrac{5\pi}{4}\right)\right) is \dfrac{\pi}{2}

 

Question 2

(i) sin^{-1}\left(\dfrac{1}{2}\right)-2~sin^{-1}\left(\dfrac{1}{\sqrt{2}}\right)

Sol :

=sin^{-1}\left[sin\left(\dfrac{\pi}{6}\right)\right]-2~sin^{-1}\left[sin\left(\dfrac{\pi}{4}\right)\right]

=\dfrac{\pi}{6}-2\times\dfrac{\pi}{4}

=\dfrac{\pi}{6}-\dfrac{\pi}{2}

=-\dfrac{\pi}{6}

 

(ii) sin^{-1}\left\{cos\left(sin^{-1}\dfrac{\sqrt{3}}{~~2}\right)\right\}

Sol :

=sin^{-1}\left\{cos\left[sin^{-1}\left(sin~\dfrac{\pi}{3}\right)\right]\right\}

=sin^{-1}\left\{cos\left(\dfrac{\pi}{3}\right)\right\}

=sin^{-1}\left(\dfrac{1}{2}\right)

=sin^{-1}\left[sin\left(\dfrac{\pi}{6}\right)\right]

=\dfrac{\pi}{6}

 

 

Question 3

Find the domain of each of the following functions:

(i) f(x)=sin^{-1}~(x^2)

Sol :

 

(ii) f(x)=sin^{-1}~x+sin~x

(iii) f(x)=sin^{-1}~\sqrt{x^2-1}

(iv) f(x)=sin^{-1}~x+sin^{-1}~2x

 

 

Question 4 

If sin^{-1}~x+sin^{-1}~y+sin^{-1}~z+sin^{-1}~t=2\pi, then find the value of x^2+y^2+z^2+t^2

 

Question 5

If (sin^{-1}~x)^2+(sin^{-1}~y)^2+(sin^{-1}~z)^2=\dfrac{3}{2}\pi^2, find the value of x^2+y^2+z^2

 

 

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